The number of solution(s) satisfying the equa | vedclass

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Question

(A) Let $f(x) = 3x^2 - 2x^3$ and $g(x) = \log_2 (x^2 + 1) - \log_2 x = \log_2 \left( \frac{x^2 + 1}{x} \right) = \log_2 \left( x + \frac{1}{x} \right)

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$1$

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(A) Let $f(x) = 3x^2 - 2x^3$ and $g(x) = \log_2 (x^2 + 1) - \log_2 x = \log_2 \left( \frac{x^2 + 1}{x} \right) = \log_2 \left( x + \frac{1}{x} \right)$.For $x > 0$,by the $AM-GM$ inequality,$x + \frac{1}{x} \geq 2$,so $g(x) = \log_2 \left( x + \frac{1}{x} \right) \geq \log_2(2) = 1$. The minimum value of $g(x)$ is $1$,occurring at $x = 1$.Now consider $f(x) = 3x^2 - 2x^3$. Its derivative is $f'(x) = 6x - 6x^2 = 6x(1 - x)$.$f'(x) > 0$ for $0  1$. Thus,$f(x)$ has a local maximum at $x = 1$.The value at the maximum is $f(1) = 3(1)^2 - 2(1)^3 = 3 - 2 = 1$.Since the maximum value of $f(x)$ is $1$ and the minimum value of $g(x)$ is $1$,the equation $f(x) = g(x)$ can only hold if $f(x) = 1$ and $g(x) = 1$ simultaneously.This occurs only at $x = 1$.Therefore,there is exactly $1$ solution.

The number of solution$(s)$ satisfying the equation $3x^2 - 2x^3 = \log_2 (x^2 + 1) - \log_2 x$ is

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